1961 AHSME Problems/Problem 40

Problem 40

Find the minimum value of $\sqrt{x^2+y^2}$ if $5x+12y=60$ .

$\textbf{(A)}\ \frac{60}{13}\qquad \textbf{(B)}\ \frac{13}{5}\qquad \textbf{(C)}\ \frac{13}{12}\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 0$

Solutions (WIP)

Solution 1

Solve for $y$ in the linear equation. $12y = 60 - 5x$ $y = 5 - \frac{5x}{12}$ Substitute $y$ in $\sqrt{x^2+y^2}$ . $\sqrt{x^2 + (5 - \frac{5x}{12})^2}$ $\sqrt{x^2 + 25 - \frac{25x}{6} + \frac{25x^2}{144}}$ $\sqrt{\frac{169x^2}{144} - \frac{25x}{6} + 25}$ To find the minimum, find the vertex of the quadratic. The x-value of the vertex is $\frac{25}{6} \cdot \frac{72}{169} = \frac{300}{169}$ . Thus, the minimum value is $\sqrt{\frac{169}{144} \cdot \frac{300^2}{169^2} - \frac{25}{6} \cdot \frac{300}{169} + 25}$ $\sqrt{\frac{10000}{16 \cdot 169} - \frac{1250}{169} + 25}$ $\sqrt{\frac{625}{169} - \frac{1250}{169} + \frac{4225}{169}}$ $\sqrt{\frac{3600}{169}}$ $\frac{60}{13}$ The answer is $\boxed{\textbf{(A)}}$ .

Solution 2

1961 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 26	Followed by Problem 28
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1961 AHSME Problems/Problem 40

Contents

Problem 40

Solutions (WIP)

Solution 1

Solution 2

See Also